Question

Prove that the points (–1, –2), (–2, –5), (–4, –6) and (–3, –3) are the vertices of a parallelogram.

Answer 1

Given: Points A(–1, –2), B(–2, –5), C(–4, –6) and D(–3, –3).

To Prove: A, B, C, D are the vertices of a parallelogram.

Proof [Step-wise]:

1. Label the points: A(–1, –2), B(–2, –5), C(–4, –6), D(–3, –3).

2. Find vectors for consecutive sides: AB = B − A 

= (–2 – (–1), –5 – (–2)) 

= (–1, –3)

BC = C – B

= (–4 – (–2), –6 – (–5)) 

= (–2, –1)

CD = D – C

= (–3 – (–4), –3 – (–6)) 

= (1, 3)

DA = A – D

= (–1 – (–3), –2 – (–3)) 

= (2, 1)

3. Compare opposite side vectors:

CD = (1, 3)

= –(–1, –3) 

= –AB

So, CD is parallel to AB and |CD| = |AB|.

DA = (2, 1)

= –(–2, –1) 

= –BC 

So, DA is parallel to BC and |DA| = |BC|.

Since each pair of opposite sides are parallel and equal in length, the quadrilateral ABCD is a parallelogram.

Alternate check: Midpoints of diagonals

AC midpoint = `((-1 + -4)/2, (-2 + -6)/2)`

= `((-5)/2, -4)` 

BD midpoint = `((-2 + -3)/2, (-5 + -3)/2)` 

= `((-5)/2, -4)` 

Diagonals bisect each other same midpoint, which is another characterization of a parallelogram.

Therefore, the points (–1, –2), (–2, –5), (–4, –6) and (–3, –3) are the vertices of a parallelogram.